Titu's lemma

The following inequality is known as Titu's lemma, Bergström's inequality, Engel's form or Sedrakyan's inequality, respectively, referring to the article About the applications of one useful inequality of Nairi Sedrakyan published in 1997, to the book Problem-solving strategies of Arthur Engel published in 1998 and to the book Mathematical Olympiad Treasures of Titu Andreescu published in 2003. It is a direct consequence of Cauchy–Bunyakovsky–Schwarz inequality. Nevertheless, in his article (1997) Sedrakyan has noticed that written in this form this inequality can be used as a mathematical proof technique and it has very useful new applications. In the book Algebraic Inequalities (Sedrakyan) are provided several generalizations of this inequality.

Statement of the inequality

For any reals [math]\displaystyle{ a_1, a_2, a_3, \ldots, a_n }[/math] and positive reals [math]\displaystyle{ b_1, b_2, b_3,\ldots, b_n, }[/math] we have [math]\displaystyle{ \frac{a^2_1}{b_1} + \frac{a^2_2}{b_2} + \cdots + \frac{a^2_n}{b_n} \geq \frac{\left(a_1 + a_2 + \cdots + a_n\right)^2}{b_1 + b_2 + \cdots + b_n}. }[/math] (Nairi Sedrakyan (1997), Arthur Engel (1998), Titu Andreescu (2003))

Probabilistic statement

Similarly to the Cauchy–Schwarz inequality, one can generalize Sedrakyan's inequality to random variable. In this formulation let [math]\displaystyle{ X }[/math] be a real random variable, and let [math]\displaystyle{ Y }[/math] be a positive random variable. X and Y need not be independent, but we assume [math]\displaystyle{ E[|X|] }[/math] and [math]\displaystyle{ E[Y] }[/math] are both defined. Then [math]\displaystyle{ \operatorname{E}[X^2/Y] \ge \operatorname{E}[|X|]^2 / \operatorname{E}[Y] \ge \operatorname{E}[X]^2 / \operatorname{E}[Y]. }[/math]

Direct applications

Example 1. Nesbitt's inequality.

For positive real numbers [math]\displaystyle{ a, b, c: }[/math] [math]\displaystyle{ \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{3}{2}. }[/math]

Example 2. International Mathematical Olympiad (IMO) 1995.

For positive real numbers [math]\displaystyle{ a,b,c }[/math], where [math]\displaystyle{ abc=1 }[/math] we have that [math]\displaystyle{ \frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \geq \frac{3}{2}. }[/math]

Example 3.

For positive real numbers [math]\displaystyle{ a,b }[/math] we have that [math]\displaystyle{ 8(a^4+b^4) \geq (a+b)^4. }[/math]

Example 4.

For positive real numbers [math]\displaystyle{ a,b,c }[/math] we have that [math]\displaystyle{ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} \geq \frac{9}{2(a+b+c)}. }[/math]

Proofs

Example 1.

Proof: Use [math]\displaystyle{ n = 3, }[/math] [math]\displaystyle{ \left(a_1, a_2, a_3\right) := (a, b, c), }[/math] and [math]\displaystyle{ \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) }[/math] to conclude: [math]\displaystyle{ \frac{a^2}{a(b + c)} + \frac{b^2}{b(c + a)} + \frac{c^2}{c(a + b)} \geq \frac{(a + b + c)^2}{a(b + c) + b(c + a) + c(a + b)} = \frac{a^2 + b^2 + c^2 + 2(ab + bc + ca)}{2(ab + bc + ca)} = \frac{a^2 + b^2 + c^2}{2(ab + bc + ca)} + 1 \geq \frac{1}{2} (1) + 1 = \frac{3}{2}. \blacksquare }[/math]

Example 2.

We have that [math]\displaystyle{ \frac{\Big(\frac{1}{a}\Big)^2}{a(b+c)} + \frac{\Big(\frac{1}{b}\Big)^2}{b(a+c)} + \frac{\Big(\frac{1}{c}\Big)^2}{c(a+b)} \geq \frac{\Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big)^2}{2(ab+bc+ac)} = \frac{ab+bc+ac}{2a^2b^2c^2} \geq \frac{3 \sqrt[3]{a^2 b^2 c^2}}{2a^2b^2c^2} = \frac{3}{2}. }[/math]

Example 3.

We have [math]\displaystyle{ \frac{a^2}{1} + \frac{b^2}{1} \geq \frac{(a + b)^2}{2} }[/math] so that [math]\displaystyle{ a^4 + b^4 = \frac{\left(a^2\right)^2}{1} + \frac{\left(b^2\right)^2}{1} \geq \frac{\left(a^2 + b^2\right)^2}{2} \geq \frac{\left(\frac{(a+b)^2}{2}\right)^2}{2} = \frac{(a + b)^4}{8}. }[/math]

Example 4.

We have that [math]\displaystyle{ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \geq \frac{(1+1+1)^2}{2(a+b+c)} = \frac{9}{2(a+b+c)}. }[/math]